I would like to know how one can calculate the area of a spherical rectangle which is defined by two longitudes and latitudes on the unit sphere.

I am well aware of answers like this question here, but I would like to do it using multidimensional integration.

My approach so far

I know I can parameterize the points on a unit sphere $$\partial\mathbb{S}^2 = \{(x,y,z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 = 1\}$$ by using spherical coordinates: $[0,\pi]\times[0,2\pi]$

$$\Omega= \begin{bmatrix} \sin\theta\cos\phi \\ \sin\theta\sin\phi \\ \cos\theta \end{bmatrix}$$

If I integrated over all of the unit sphere's area, I would do the following: $$\int_F do = \int_0 ^\pi\int_0 ^{2\pi}|\partial_{\theta}\Omega\times \partial_{\phi}\Omega|d\phi d\theta$$

Now, however, I do not need to integrate over the whole unit sphere so I must change my area of integration. Furthermore, I believe that I would have to change my parametrization slightly.

Let's say the rectangle is $b$ high (distance between two latitudes), and $c$ wide, (the distance between two latitudes) as well as $a$ above the equator.

Since $\phi$ "is symmetrical" instead of integrating from $[0,2\pi]$ we can integrate from $[0,c]$, (right?), but how do I integrate over $\theta$, since not only the height of the rectangle is important but also how far away it is from the equator.

Your help is greatly appreciated.

sphere (Sorry for the bad picture)

  • $\begingroup$ I must be missing something here. The area of the segment (the band going round the sphere completely between two given latitudes) is $2 \pi \cdot h \cdot R$, $h$ being the height of the band ($h = R \cdot (\sin(\phi_2) - \sin(\phi_1))$) as shown e. g. in en.wikipedia.org/wiki/Spherical_segment . Then we only have to take a linear part of that based on the longitudes to get the result: $A = (\theta_2 - \theta_1) \cdot (\sin(\phi_2) - \sin(\phi_1)) \cdot R^2$ Why the complexity with integrals in this case? $\endgroup$ – Alfe Jul 18 '18 at 15:16

The kind of spherical coordinates you are using can be described as a longitude $\phi$ and a co-latitude $\theta.$ Unlike the latitude of mapmakers, which is zero at the equator and has a maximum magnitude of $90$ degrees ($\frac\pi2$ radians) north or south, the co-latitude is zero at the north pole of your sphere (at Cartesian coordinates $(0,0,1)$) and is $\pi$ radians at the other pole; it is $\frac\pi2$ radians at the equator.

Since you want one boundary of your integral to be at an angle $a$ "above" the equator—that is, closer to the $\theta=0$ pole—the co-latitude of that boundary is $\frac\pi2 - a.$ The co-latitude of the other boundary is another $b$ radians closer to the pole, so it is $\frac\pi2 - a - b.$

Note that the bound at $\frac\pi2 - a - b$ has a smaller co-latitude, so the integration interval "starts" there rather than "ending" there. We therefore have the integral $$ \int_{\frac\pi2 - a - b}^{\frac\pi2 - a} \,\int_0^c \left\lvert\partial_{\theta}\Omega\times \partial_{\phi}\Omega \right\rvert \,d\phi \,d\theta. $$

By symmetry of the figures above and below the equator, we will get the same result if we integrate $$ \int_{\frac\pi2 + a}^{\frac\pi2 + a + b} \!\int_0^c \left\lvert\partial_{\theta}\Omega\times \partial_{\phi}\Omega \right\rvert \,d\phi \,d\theta. $$

  • $\begingroup$ Since $ \left\lvert\partial_{\theta}\Omega\times \partial_{\phi}\Omega \right\rvert = \sin(\phi)$. So $\int_{\frac\pi2 + a}^{\frac\pi2 + a + b} \!\int_0^c \left\lvert\partial_{\theta}\Omega\times \partial_{\phi}\Omega \right\rvert \,d\phi \,d\theta. = \int_{\frac\pi2 + a}^{\frac\pi2 + a + b} \!\int_0^c \sin(\phi) \,d\phi \,d\theta. = \int_{\frac{\pi}{2} +a}^{\frac{\pi}{2} +a+b} (1-\cos(c))d\theta = [1-\cos(c)]_{\frac{\pi}{2} +a} ^{\frac{\pi}{2} +a+b} = b(\cos(c)-1)$ Is that correct. In that case the area would not depend on the distance from the equator which seems a bit odd. $\endgroup$ – user521723 Jan 22 '18 at 13:08
  • $\begingroup$ Oops. I meant $= b(1-\cos(c))$ $\endgroup$ – user521723 Jan 22 '18 at 13:17
  • $\begingroup$ I thought $\left\lvert\partial_\theta\Omega\times\partial_\phi\Omega\right\rvert = \sin(\theta).$ The result should be proportional to $c.$ A possibly surprising fact is that the result is also proportional to the distance between the parallel planes on which the two "lines" of latitude lie. $\endgroup$ – David K Jan 22 '18 at 15:06
  • $\begingroup$ Yes of course. You are right. My fault. Thanks for clarifying $\endgroup$ – user521723 Jan 22 '18 at 15:53

Don't you worry, the picture is clear enough. There are other issues: when we speak of spherical geometry, we have in mind objects built of arcs of great circles. The question you linked to is about that. And as it's said, there, you can't have such a rectangle with four right angles (because, if the inner angles of such a rectangle are $\alpha$, $\beta$, $\gamma$ and $\delta$, and the radius is $1$, the area of the rectangle would be the spherical excess, i.e. $\alpha+\beta+\gamma+\delta-2\pi$). In your case, you do have four right angles, but that's ok, because circles of equal latitude are not great circles. You're proposing spherical coordinates, and that's fine and natural, because longitudes and latitudes are angles. So I'll assume that the $a, b, c$ in your question are angles. Then, the integration limits are pretty clear: $0$ to $c$ for $\phi$ (as you wrote), and $a$ to $a+b$ for $\theta$. I guess you'll understand that you'll have to make your $|\partial_{\theta}\Omega\times \partial_{\phi}\Omega|$ a bit more explicit to calculate something, and that (let's be honest) would be reason enough for me not to use an integral, because (much like Archimedes) I don't even need a pencil, otherwise: $c\,(\cos(a)-\cos(a+b))$. The reasoning is here, and I think it's an excellent example of human genius. Of course, if we make the expression mentioned above more explicit, the calculation won't be any more complicated, but Archimedes had to do without that convenient formalism.


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